# Math 2280 - Lecture 25 Dylan Zwick Fall 2013

```Math 2280 - Lecture 25
Dylan Zwick
Fall 2013
So far in our study of how to solve systems of ODEs we’ve focused
almost exclusively on systems of the form:
x′ = Ax.
This is just the homogeneous case. It’s an extremely important case,
but it can also be viewed as, in some sense, the easiest possible case! In
general, we’d like to be able to solve a system of ODEs of the form:
x′ = Ax + f(t)
for any value of of the function f(t). How do we do this? Well, the same
way we did it in chapter 3. We guess! In particular, if our function f(t) is
a combination (sums and products) of polynomials, exponentials, cosines,
and sines, then we use the matrix analogue of the method of undetermined
coefficients. There’s also a matrix analogue of the method of variation of
parameters. We’ll be discussing both today.
Today’s lecture corresponds with section 5.6 from the textbook, and the
assigned problems are:
Section 5.6 - 1, 6, 10, 17, 19
1
The Method of Undetermined Coefficients
The method of undetermined coefficients for a system of ODEs works
pretty much the way it did for the ODEs we studied in chapter 3. It’s
best understood in the context of examples.
Example - Find a particular solution to the system of ODEs:
′
x =
3 4
3 2
2
x+
0
t2
The only issue with this method is when our function f(t) exhibits duplication with our homogeneous solution. So, for example, if we have the
system:
′
x =
4 2
3 −1
x−
15
4
te−2t
then out homogeneous solution is:
xh (t) = c1
1
−3
−2t
e
+ c2
2
1
e5t
which exhibits duplication with our preliminary trial solution xp (t) =
ate−2t + be−2t . Now, this is where this method differs from the earlier
method of undetermined coefficients. Here, we need to multiply by t until
we no longer exhibit duplication, but we still have to include all lower
terms! So, our new trial solution would be:
xp (t) = at2 e−2t + bte−2t + ce−2t .1
We could solve this, but it turns out that the method of variation of
parameters is actually simpler to use in this case. So, we’ll postpone the
The Method of Variation of Parameters
In matrix form the method of variation of parameters is actually pretty
easy to formulate.
Suppose we have a system of ODEs:
x′ = P(t)x + f(t)
1
This is because we could have two linearly independent vectors c1 e−2t and c2 e−2t , a
situation that didn’t come up in chapter 3.
3
and we already have a complete set of solutions to the associated homogeneous system. From these solutions we can construct a fundamental
matrix Φ(t). We’re going to try to write our particular solution as a linear
combination of our linearly independent solutions to the homogeneous
equation, but instead of being a linear combination involving constant coefficients, the coefficients will be functions. In matrix notation this is:
xp (t) = Φ(t)u(t).
If we differentiate this we get:
x′p (t) = Φ(t)′ u(t) + Φ(t)u′ (t).
Now, if we use the relations:
x′p (t) = P(t)xp + f(t)
and
Φ′ (t) = P(t)Φ(t)
then plugging these relations into our above system and simplifying
we get:
f(t) = Φ(t)u′ (t).
So, if we solve this for u(t) we get:
u(t) =
Z
Φ(t)−1 f(t)
and therefore
xp (t) = Φ(t)
Z
4
Φ(t)−1 f(t).
If P(t) = A, where A is a constant matrix, then we have the fundamental matrix Φ(t) = eAt , and our particular solution given by:
At
xp (t) = e
Z
e−At f(t).
To solve a given initial value problem we get:
At
At
x(t) = e x0 + e
Z
t
e−As f(s)ds.
0
Note the striking similarity between this equation and the corresponding equation from section 1.5.
Example - Find the general solution to the system:
′
x =
4 2
3 −1
x−
5
15
4
te−2t .
Room for the example problem.
6
Notes on Homework Problems
Problems 5.6.1, 5.6.6, and 5.6.10 are, increasingly more difficult, problems
where you have to “guess and check” a particular solutions given the form
of f(t). Not too difficult, although honestly a bit tedious.
Problems 5.6.17 and 5.6.19 are exercises using variation of parameters.
Should be straightforward, although again kind of involved and, honestly,
a bit tedious. But good for you none the less! :)
7
```